Making Craters With Rays
Notes to Accompany The Video Clip
By Dr. Larry Krumenaker
Copyright 2009 by The Classroom Astronomer Magazine
All Rights Reserved
Contact at firstname.lastname@example.org .
This four second clip shows a glass marble being dropped into a 3 inch deep pan
of flour covered with a thin, uniform layer of cocoa powder to simulate a lunar surface.
The ball appears as a blur on the top edge of the film at about the 2 second mark. When
the ball hits the ‘lunar surface,’ it creates a crater and simultaneously ejects the matter
from the crater hole. This produces a set of three prominent white streaks or rays
emanating from the hole, similar to what astronomers see in craters on the moon that are
presumed to be produced in the more recent geological past.
This clip can also be used for some physics and geometry problems.
The first possible exercise is estimating the speed of the marble (meteor, when it
was in motion, meteorite afterwards) and the height from which it was dropped.
One must first create a scale for the image, which will depend on the way the
video is viewed. The pan is 3 inches (7.5 cm deep) and from outer edge to outer edge it
is 22cm. For ease, the ball was dropped in the approximate center of the pan. The video
shows 20 frames per second and the ball takes but one frame (0.05 seconds) to go from
film edge to impact.
However, the film was shot at an angle, which is somewhere between 45 and 60
degrees from the horizontal. This foreshortens the visible track of the impacting ball by a
factor equal to the cosine of the angle from the horizontal. Thus the actual track length is
the measured amount (coincidentally about one-half the pan size or 11cm after using a
scale for your actual viewing size) divided by the cosine of the angle. The cosine of 45
degrees is .707 and for 60 degrees the value is .866 . The actual path seen is thus 15.5 to
19 cm long. One can use the average of 17.3 cm with a range of plus-minus 1.7 cm.
Since the average speed during this interval is Δv/Δt, this works out to be
17.3/0.05 or 346 cm per second, plus or minus 34 cm/sec. This equal the final velocity
V f at the moment of initial impact.
From what height was this ball released?
The equations needed are V f = gt (the gravitational constant g or 980 cm/sec 2
times time in seconds from the moment of release) and height = ½ g t 2 .
Using V f from above, we get t = 0.353 seconds have elapsed from release to
impact. Substituting this into the height equation we get a height of 61cm. This is about
two feet high or 60 cm. However, with our uncertainty, we could have released the ball
from as high as 73.6 cm or 29 inches to 18 inches. After the fact, we recreated the
experiment and found it to be roughly 30 inches.
Given the Vf above, we could calculate the kinetic energy (KE) upon impact as ½
m Vf 2 and use the change in KE to find the change in Potential Energy (PE) and the
height again. The values should be identical. The mass is approximately 0.75 ounces or
21 grams. The KE would thus be about 0.01 joules.
The ball itself is 2.0 cm in diameter. It clearly comes to rest below the surface
level, though still partially visible in the hole. It can be estimated to have gone, then, only
about half way down into the flour, or about 3 cm for the ball’s lower edge up from the
pan. The ball’s center would have thus penetrated 2cm. One can use these values to
calculate a deceleration value, given that it took only about 0.05 seconds to come to a
stop. Acceleration would be a negative 6.92 meters per second per second, after
Determining the exact angle of the camera
The sides of the tray going away from the camera will appear shorter by the sine
of the angle from the horizontal. From above all the sides would have the same length on
the screen. From horizontal, the sides moving away would be so foreshortened as to have
zero length. Thus, measuring the ratio of the sides moving away versus the length side to
side of the tray would give us the sine of the angle of the camera above being horizontal.
Using the arc-sin function would get that angle itself.
On our screen, we measured 15.5 cm for the sides lining away and 23 cm for the
nearer edge. Knowing that they should be 22 cm each, and that we have about a 0.5 cm
uncertainty, given the irregularity of the edges and the curved corners, we estimated a
value of 43 degrees from the horizontal.